Perms and Combs Question (1 Viewer)

Ambility · Oct 7, 2015

Cambridge 3 Unit HSC 10F, 16b.

$Ten coloured marbles are placed in a row. What is the minimum number of colours needed to guarantee at least 10 000 different patterns? [HINT: This will need a guess-and-check approach.]$\\$My logic is: $3^{10}=59049$ but $2^{10}=1024$ therefore 3 colours are required. Cambridge answers say 4 colours are required.$

InteGrand · Oct 7, 2015

Ambility said:
Cambridge 3 Unit HSC 10F, 16b.

$Ten coloured marbles are placed in a row. What is the minimum number of colours needed to guarantee at least 10 000 different patterns? [HINT: This will need a guess-and-check approach.]$\\$My logic is: $3^{10}=59049$ but $2^{10}=1024$ therefore 3 colours are required. Cambridge answers say 4 colours are required.$

We are meant to assume the ten marbles are given to us. We can't choose from colours however many we like, we assume ten marbles are given with fixed numbers of each colour. E.g. with two colours, we could be given 1 W and 9 B (White and Black), or 2 W and 8 B, etc. We need to see whether for any of these GIVEN combinations of the colours, it is possible to obtain 10000 or more different patterns. For example, with 1 W and 9 B, the number of different possible patterns is just 10. With 2 W and 8 B, it is (10!)/(2! • 8!) = 45. We can keep checking with different FIXED combinations of 2-colour sets of 10 marbles and we will see that we never get 10000 different pattern.

So we can try with a FIXED set of 3. One possibility is 3 R, 4 G, 3 B (Red, Green, Blue), etc. Then show no matter what set of 10 marbles given, if there's only 3 colours, there can never be 10000 possibilities.

Drsoccerball · Oct 7, 2015

$$ List out possible numbers $ 1,2,3,4,...,10$

$$ The amount of arrangements is $ = \frac{10!}{r!} = 10000$

$r is the amount of colours which are repeated $ \therefore 10-r = $ Minimum amount of different coloured balls required$

$r! =362.88 \therefore $ r is closest to $ 6! $ As $ 5! $ wouldn't provide the $ 10000.$

$\therefore 10-6=4$

InteGrand · Oct 7, 2015

InteGrand · Oct 7, 2015

Drsoccerball said:
$$ List out possible numbers $ 1,2,3,4,...,10$

$$ The amount of arrangements is $ = \frac{10!}{r!} = 10000$

$r is the amount of colours which are repeated $ \therefore 10-r = $ Minimum amount of different coloured balls required$

$r! =362.88 \therefore $ r is closest to $ 6! $ As $ 5! $ wouldn't provide the $ 10000.$

$\therefore 10-6=4$

$What do you mean ``amount of colours which are repeated ''? Would you say $r=5$ in this case?: $G,G,R,R,R,$ random other single colours (red, green, blue for $R,G,B$). It wouldn't be right to say that the number of possibilities is $\frac{10!}{5!}$, it needs to be $\frac{10!}{2!\cdot 3!}$ here, so we can't use that method.$

Drsoccerball · Oct 7, 2015

InteGrand said:
$What do you mean ``amount of colours which are repeated ''? Would you say $r=5$ in this case?: $G,G,R,R,R,$ random other single colours (red, green, blue for $R,G,B$). It wouldn't be right to say that the number of possibilities is $\frac{10!}{5!}$, it needs to be $\frac{10!}{2!\cdot 3!}$ here, so we can't use that method.$

No but I am making the assumption that there is only one repeated colour. It would be silly to increase the amount of combinations by adding the same colour.

Drsoccerball · Oct 7, 2015

My bad my method doesn't even give you 10000 D:

turnerloos · Oct 7, 2015

Drsoccerball said:
My bad my method doesn't even give you 10000 D:

tskkk.. not even state ranking material bro.

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Perms and Combs Question (1 Viewer)

Ambility

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InteGrand

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Drsoccerball

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InteGrand

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turnerloos

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